User jkun - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:33:10Z http://mathoverflow.net/feeds/user/6429 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65881/equivalent-conditions-for-cohopfian-groups Equivalent conditions for coHopfian groups jkun 2011-05-24T17:32:31Z 2012-10-28T17:05:33Z <p>A lot of work has been done on determining whether particular classes of groups are coHopfian, in particular the (sufficiently large) braid groups $B_n$ modulo center, and certain classes of 3-manifold groups.</p> <p>Are there any equivalent conditions for a group to be coHopfian? I'm aware this is a big open problem, so I'm specifically looking for established results or necessary conditions.</p> <p>Of course, sufficient conditions are helpful as well, but most of what I've read in terms of sufficient conditions is "$G$ can be realized as a member of a class X, which we prove is coHopfian," and this is not what I want.</p> http://mathoverflow.net/questions/68006/kolmogorov-complexity-and-proof-techniques/93015#93015 Answer by jkun for Kolmogorov Complexity and Proof Techniques jkun 2012-04-03T16:25:33Z 2012-04-03T16:25:33Z <p>Using the same technique, one can construct infinitely many statements which are true with probability arbitrarily close to 1, but are nonetheless unprovable. See lemma 4 in <a href="http://theory.stanford.edu/~trevisan/cs172/notek.pdf" rel="nofollow">http://theory.stanford.edu/~trevisan/cs172/notek.pdf</a></p> http://mathoverflow.net/questions/78410/finitely-presented-infinite-group-with-no-element-of-infinite-order Finitely presented infinite group with no element of infinite order? jkun 2011-10-18T02:15:11Z 2011-10-19T03:53:35Z <p>Is there an example of a finitely presented infinite group in which every element has finite order? Or, is it known that every finitely presented infinite group has an element of infinite order?</p> <p>I asked this question of math.stackexchange, thinking it might be trivial (for finitely generated groups there are numerous counterexamples...), but it seems the question is wide open. So more specifically, could someone point me to partial results, or give a good reason why this problem won't be solved any time soon?</p> http://mathoverflow.net/questions/8846/proofs-without-words/69756#69756 Answer by jkun for Proofs without words jkun 2011-07-07T23:25:37Z 2011-07-07T23:25:37Z <p>Another proof of the sum of the first $n$ squares, relying on the knowledge of the formula for the sum of the first $n$ numbers:</p> <p>$1^2 + 2^2 + \dots + n^2 = n(n+1)(2n+1)/6$</p> <p><img src="http://jeremykun.files.wordpress.com/2011/06/triangle-proof.png" alt="alt text"></p> <p>This one has a similar flavor to the fabled proof by Gauss of the sum of the first $n$ numbers. It's a good follow up for students after Gauss's proof.</p> http://mathoverflow.net/questions/8846/proofs-without-words/69594#69594 Answer by jkun for Proofs without words jkun 2011-07-06T00:39:00Z 2011-07-06T00:39:00Z <p>Can you tile an 8x8 chessboard with one corner cut off with dominoes of dimension 3x1?</p> <p><img src="http://jeremykun.files.wordpress.com/2011/06/chessboard-1-by-3.jpg" alt="alt text"></p> <p>This is a simple way to show that choosing a useful coloring can make a proof trivial.</p> <p>This proof was also a result of the Conjecture and Proof class in the Budapest Semesters in Mathematics. It was one of the first problems encountered there, hence not <em>that</em> hard :)</p> http://mathoverflow.net/questions/62185/freiheitssatz-implies-a-finitely-generated-one-relator-group-embeds-in-a-two-gene Freiheitssatz implies a finitely generated one relator group embeds in a two-generator one relator group? jkun 2011-04-19T01:09:00Z 2011-04-19T19:44:32Z <p>I've read that every finitely generated one relator group embeds in a two generator one relator group, and that this fact follows from the Freiheitssatz. </p> <p>Unfortunately, the only proof I can find of this fact applies <a href="http://www.jstor.org/stable/91573" rel="nofollow">B.H. Neumann's proof</a> for denumerable n-relator groups, and it doesn't seem to use the Freiheitssatz. Further, I haven't found any mention of this in Lyndon and Schupp, but it's possible I overlooked a more general theorem from which this follows.</p> <p>My question is: does this fact truly follow from the Freiheitssatz? Is the proof trivial? I apologize if it is; unfortunately I am new to one relator groups.</p> http://mathoverflow.net/questions/61759/does-the-group-given-by-this-presentation-have-an-element-of-order-2 Does the group given by this presentation have an element of order 2? jkun 2011-04-14T22:26:52Z 2011-04-14T22:36:34Z <p>Suppose $G$ has the presentation $\langle t, x_1, x_2, ... | R \rangle$ where each relator in $R$ has the form $t^{-1}x_it = x_j$ for some $i,j$. Does $G$ have an element of order 2?</p> <p>This is an HNN extension of a free group, if that changes anything.</p> http://mathoverflow.net/questions/58035/sufficient-conditions-for-free-indecomposability Sufficient Conditions for Free Indecomposability jkun 2011-03-10T04:23:06Z 2011-03-11T22:10:58Z <p>An interesting fact was relayed to me in <a href="http://mathoverflow.net/questions/57902/fundamental-groups-of-closed-hyperbolic-3-manifolds-are-freely-indecomposableBlockquote" rel="nofollow">another question of mine</a> that </p> <p><Blockquote> If $M$ is any closed manifold with universal cover homeomorphic to $R^n$ for $n>1$ then $\pi_1(M)$ is freely indecomposable. </Blockquote></p> <p>What are some other sufficient conditions for the free-indecomposability of a group? Are there any interesting necessary conditions?</p> http://mathoverflow.net/questions/57902/fundamental-groups-of-closed-hyperbolic-3-manifolds-are-freely-indecomposable Fundamental groups of closed hyperbolic 3-manifolds are freely indecomposable jkun 2011-03-09T02:00:18Z 2011-03-09T04:33:20Z <p>I believe the following statement is true, and I've even seen it referenced <a href="http://mathoverflow.net/questions/31904/closed-3-manifolds-with-free-abelian-fundamental-groups" rel="nofollow">here</a>. Could someone point me to a proof?</p> <blockquote> <p>The fundamental group of a closed hyperbolic 3-manifold is not a free product.</p> </blockquote> http://mathoverflow.net/questions/78410/finitely-presented-infinite-group-with-no-element-of-infinite-order Comment by jkun jkun 2011-10-18T03:01:47Z 2011-10-18T03:01:47Z So the von Neumann problem talks about non-amenable groups without non-abelian free subgroups, but I'm just looking for the existence of a copy of $\mathbb{Z}$. Wouldn't this be easier? http://mathoverflow.net/questions/60160/decidability-in-groups Comment by jkun jkun 2011-10-18T02:20:43Z 2011-10-18T02:20:43Z There is active research going on about whether such problems are decidable if we assume the word problem has a solution. For instance, determining whether a group is a 3-manifold group is decidable, given a presentation and a solution to the word problem. http://mathoverflow.net/questions/62185/freiheitssatz-implies-a-finitely-generated-one-relator-group-embeds-in-a-two-gene/62189#62189 Comment by jkun jkun 2011-04-19T22:23:14Z 2011-04-19T22:23:14Z This is a very interesting proof! Unfortunately I am somewhat bound to use the Freiheitssatz in my exposition, but these stronger results are always interesting to note. Thank you! http://mathoverflow.net/questions/62185/freiheitssatz-implies-a-finitely-generated-one-relator-group-embeds-in-a-two-gene/62195#62195 Comment by jkun jkun 2011-04-19T18:49:13Z 2011-04-19T18:49:13Z I think I need a slight clarification, because I was promised this works for every one relator group, regardless of the relator. I know the Freiheitssatz extends to the case when the relator does not necessarily involve all generators, but the chosen subset still excludes a generator involved in $r$. However I'm not so sure this proof extends in a similar way. Take the case where $r = g_1^k$ involves a single generator, then it seems to me that any HNN extension (at least, the construction used here) would necessarily have at least three generators. Is there something I'm missing here? http://mathoverflow.net/questions/61759/does-the-group-given-by-this-presentation-have-an-element-of-order-2/61760#61760 Comment by jkun jkun 2011-04-16T16:26:57Z 2011-04-16T16:26:57Z No. It is definitely an HNN extension of a free group. I just posed the more general question because I didn't know of this theorem on HNN extensions, and thought to answer it from a different angle. http://mathoverflow.net/questions/61759/does-the-group-given-by-this-presentation-have-an-element-of-order-2/61760#61760 Comment by jkun jkun 2011-04-16T03:31:01Z 2011-04-16T03:31:01Z This is unfortunate for at least one author who I'll leave anonymous, because it provides a counterexample to a theorem in one of his books on one relator groups. http://mathoverflow.net/questions/58035/sufficient-conditions-for-free-indecomposability/58038#58038 Comment by jkun jkun 2011-03-11T03:30:58Z 2011-03-11T03:30:58Z I apologize for the ignorance, but is that a variation/application of Helly's theorem on convex sets? Could you point me to a proof of the sufficient condition above? http://mathoverflow.net/questions/57902/fundamental-groups-of-closed-hyperbolic-3-manifolds-are-freely-indecomposable/57907#57907 Comment by jkun jkun 2011-03-09T22:27:48Z 2011-03-09T22:27:48Z This is the proof that was communicated orally to me, but I was looking for the name or to see it in a book somewhere. This is just what I needed. Thanks! http://mathoverflow.net/questions/26420/beginner-group-theory-question Comment by jkun jkun 2010-05-30T05:36:30Z 2010-05-30T05:36:30Z my mistake. will read the faq next time.